Spontaneity, Entropy and Free Energy

Dr. Lu Yunpeng yplu@ntu.edu.sg

Office: CBC-06-21 Tel: 6513-2747

Consider the following reaction at equilibrium:

k [B] j [A]

m [D] l [C] Kc =

The Equilibrium Constant, Kc

Square brackets = concentrations of species at equilibrium.

j, k, l, and m = coefficients in the balanced equation.

K = equilibrium constant (given without units).

jA + kB lC + mD

Equilibrium Expressions Involving Pressure, Kp

N2(g) + 3H2(g) 2NH3(g)

( )( )( )

3

2 2

2

NH

p 3

N H

P =

P PK

[ ][ ][ ]

23

32 2

NH = N H

K

• Kc involves concentrations. • Kp involves pressures.

Fritz Haber (1868-1934) Nobel Laureate in Chemistry, 1918

The Reaction Quotient, Q

k [B] j [A]

m [D] l [C] Q =

In general, ALL reacting chemical systems at any time are characterized by their REACTION QUOTIENT, Q.

Square brackets = concentrations of species at the time of measurement.

j, k, l, and m = coefficients in the balanced equation.

Q = reaction quotient (given without units).

The most valuable time to calculate Q is apparently at the beginning of reactions. It tells about the change direction!

Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous.

Chemical Thermodynamics

• Describe chemical reactions from an energy point of view:

-At equilibrium (relation with K?) -At non-equilibrium (relation with Q?) • Predict if a reaction is favourable and spontaneous

(will the reaction occur without external help?) • Predict if a molecule is stable

• First law of thermodynamics: The law of conservation of energy; energy cannot be created or destroyed.

• State Function: Quantity in which its determination is path independent.

• ∆U = q + w: The change in internal energy of a system is a function of heat and work done on or by the system.

• ∆H: Heat transferred at constant pressure. • Exothermic Process: ∆H < 0 • Endothermic Process: ∆H > 0

What you should have known about Thermodynamics

Spontaneous Processes A process that does occur under a specific set of conditions is called a spontaneous process. A process that does not occur under a specific set of conditions is called nonspontaneous.

Spontaneous Process Nonspontaneous Process

Ice melting at room temperature

Sodium metal reacts with water

A ball falling downhill

Iron rusting at wet condition

Water freezing at -10˚C

Spontaneous changes occur only in the direction that leads to equilibrium. Systems never change spontaneously in a direction that takes them farther from equilibrium.

A process that results in a decrease in the energy of a system often is spontaneous:

The sign of ΔH alone is insufficient to predict spontaneity in every circumstance!

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH° = -890.4 kJ/mol

H2O(l) H2O(s) T > 0°C; ΔH° = -6.01 kJ/mol

Driving force in spontanenous processes

Solvation of a Soluble salt: NH4NO3(s) dissolves spontaneously even the process is endothermic. (∆H > 0) Expansion of a gas in the vacuum is energy neutral. (∆H = 0) More?

Entropy, S • An index or measurement of energy dispersal to microstates.

• The driving force for a spontaneous process is an increase in the entropy of the universe.(Thermodynamics Second Law)

Rudolf Julius Emanuel Clausius ( 1822 – 1888)

• Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state.

• Nature spontaneously proceeds toward the states that have the highest probabilities of existing.

The Microstates That Give a Particular Arrangment (State): 4 particles

Positional Entropy • A gas expands into a vacuum because the

expanded state has the highest positional probability of states available to the system

• .Therefore: Ssolid < Sliquid << Sgas

Which of the following substances has the higher positional probability per mole?

a) Liquid water at 100°C. b) Gaseous water at 100°C. c) Because the

temperatures are the same, the positional probabilities are the same.

Second Law of Thermodynamics • In any spontaneous process there is always an

increase in the entropy of the universe. • The entropy of the universe is increasing.

ΔSuniverse = ΔSsystem + ΔSsurroundings

ΔSuniverse > 0 for a spontaneous process ΔSuniverse < 0 for a nonspontaneous process ΔSuniverse = 0 for an equilibrium process

Note: S and ΔS, without any subscript, refer to system’s entropy and entropy change!

Interplay of ∆Ssys and ∆Ssurr in Determining the Sign of ∆Suniv

For any spontaneous process, a) ∆Suniv, ∆Ssys, and ∆Ssurr

must all be positive. b) ∆Suniv and ∆Ssys must be

positive. c) ∆Ssys and ∆Ssurr must be

positive. d) Only ∆Suniv must be

positive. e) Only ∆Ssys must be

positive.

• The sign of ΔSsurr depends on the direction of the heat flow.

• The magnitude of ΔSsurr depends on the temperature.

The Effect of Temperature on Spontaneity

H2O(l) H2O(g)

surr = ∆∆ −

HST

at constant T and P

For which process is ∆S positive?

a) Water freezing at -20°C b) An ideal gas being

compressed reversibly at a constant temperature and pressure

c) A precipitation reaction d) A spontaneous

endothermic process at a constant temperature and pressure

Gibbs Free Energy (G) Measurements on the surroundings are seldom made, limiting the use of the second law of thermodynamics. Gibbs free energy (G) or simply free energy can be used to express spontaneity more directly. The change in free energy for a system is (at constant T and P):

G = H – TS

ΔG = ΔH – TΔS

Josiah Willard Gibbs (1839 –1903)

Free Energy Change (ΔG)

• A process (at constant T and P) is spontaneous in the direction in which the free energy decreases. Negative ΔG means positive ΔSuniv.

univ = (at constant and )∆∆ −

GS T PT

ΔG = ΔH – TΔS

For the vaporization of a liquid at a given pressure,

a) ∆G is positive at all temperatures.

b) ∆G is negative at all temperatures.

c) ∆G is positive at low temperatures, but negative at high temperatures.

d) ∆G is negative at low temperatures, but positive at high temperatures.

A liquid is vaporized at its boiling point. Predict the signs of: W q ∆H ∆S ∆Ssurr ∆G Explain your answers.

In which case must a reaction be spontaneous at all temperatures?

a) ∆H is positive, and ∆S is positive.

b) ∆H is negative, and ∆S is negative.

c) ∆H is positive, and ∆S is negative.

d) ∆H is negative, and ∆S is positive.

e) ∆H = 0, and ∆S = 0.

Interplay of ∆H and ∆S for a Process and the Resulting Dependence of Spontaneity on Temperature

Entropy Changes in Chemical Reaction

• The entropy of a perfect crystal at 0 K is zero.

• The entropy of a substance increases with temperature.

The third law of thermodynamics

N2(g) + 3H2(g) → 2NH3(g)

• Fewer molecules means fewer possible configurations, the decrease of entropy.

• For reaction involved gaseous molecules, entropy change has the same sign as molecular numbers change.

4NH3(g) + 5O2(g) → 4NO (g) + 6H2O(g) CaCO3(s) → CaO(s) + CO2(g) 2SO2(g) + O2(g) → 2SO3(g)

Entropy Changes in Chemical Reaction ─── Gaseous Molecules

For the dissociation reaction of the acid HF, ∆S is negative. HF(aq) H+(aq) + F–(aq) Which statement best explains why ∆S is negative?

a) Each HF molecule produces two ions when it dissociates.

b) The ions are hydrated. c) The reaction is expected to

be exothermic, so ∆S should be negative.

d) The reaction is expected to be endothermic, so ∆S should be negative.

Entropy Changes in Chemical Reaction

Standard Molar Entropy Values (S°)

• Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.

ΔS°reaction = ΣnpS°products – ΣnrS°reactants

Calculate ΔS° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: S° (J/K·mol) Na(s) 51 H2O(l) 70 NaOH(aq) 50 H2(g) 131

Standard Free Energy Change (ΔG°)

• The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.

ΔG° = ΔH° – TΔS°

ΔG°reaction = ΣnpG°products – ΣnrG°reactants

The standard free energy of reaction (ΔG°reaction) is free-energy change for a reaction when it occurs under standard-state conditions.

The following conditions define the standard states of pure substances and solutions are:

Standard Free Energy Change (ΔG°)

Matter Standard States Gas 1 atm pressure

Liquid Pure liquid Solid Pure solid

Element the most stable allotropic form at 1 atm and 25°C

Solution 1 molar concentration

Consider the following system at equilibrium at 25°C. PCl3(g) + Cl2(g) PCl5(g)

∆G° = −92.50 kJ

What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain.

Standard Free Energy Change of Formation, ΔGf° • The standard free energy of formation, ΔGf°

–The change in free energy for a reaction in which 1 mol of a substance in its standard state is formed from its elements in reference forms in their standard states. i.e. 6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s) And ΔGf° for an element in its standard state is zero.

ΔG° = [ Σnp ΔGf°(products) - Σnr ΔGf°(reactants)]

For a chemical reaction,

In tables of thermodynamic data provided in chemistry books, one finds ∆H°f, ∆G°f and S° listed. Briefly, explain why the entropy data are supplied as S°, while the enthalpy and free energy data are in the form of ∆H°f and ∆G°f, respectively.

The Dependence of Free Energy on Pressure It is the sign of ΔG (not ΔG°) that determines spontaneity of real reaction. The relationship between ΔG and ΔG° is:

R is the gas constant (8.314 J/K·mol).

T is the kelvin temperature.

Q is the reaction quotient.

ΔG = ΔG° + RT lnQ

The equilibrium constant, KP, for the reaction N2O4(g) ⇌ 2NO2(g)

is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.4 kJ/mol. In a certain experiment, the initial pressures are PN2O4

= 0.453 atm and PNO2 = 0.122 atm.

Calculate ΔG for the reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously to establish equilibrium.

(298 K)(ln 0.0329) = 8.314×10-3 kJ

K∙mol 5.4 kJ mol

+

= 5.4 kJ/mol – 8.46 kJ/mol

ΔG = ΔG° + RT lnQ

= –3.1 kJ/mol

QP = (0.122)2

0.453 (PNO2)2

PN2O4 = = 0.0329

Free Energy and Equilibrium

At equilibrium, ΔG = 0 and Q = K:

0 = ΔG° + RT ln K ΔG° = –RT ln K

At equilibrium, ΔG = 0 and Q = K:

0 = ΔG° + RT ln K

Free Energy and Equilibrium

ΔG° = –RT ln K

At constant pressure, the reaction 2NO2(g) → N2O4(g) is exothermic. The reaction (as written) is

a) always spontaneous. b) spontaneous at low

temperatures, but not at high temperatures.

c) spontaneous at high temperatures, but not at low temperatures.

d) never spontaneous.

A mixture of hydrogen and chlorine remains unreacted until it is exposed to ultraviolet light from a burning magnesium strip. Then the following reaction occurs very rapidly: H2(g) + Cl2(g) → 2HCl(g) ∆G = –45.54 kJ ∆H = –44.12 kJ ∆S = –4.76 J/K Select the statement that best explains this behavior. 1. The reactants are thermodynamically

more stable than the products. 2. The reaction has a small equilibrium

constant. 3. The ultraviolet light raises the

temperature of the system and makes the reaction more favorable.

4. The negative value for ∆S slows down the reaction.

5. The reaction is spontaneous, but the reactants are kinetically stable.

More Calculation Examples

Problem: Calculate entropy change for the surrounding in the reaction in the standard state: 2H2(g) + O2(g) → 2H2O (l)

surr = ∆∆ −

HST

ΔH° = ΣnproductsΔHf° products – ΣnreactantsΔHf° reactants

H2 (g) O2(g) H2O(g) ΔHfº (kJ/mol) 0. 0. -285.85

ΔHº = 2 x (-285.85) – 2 x (0.0) – 2 x (0.0) = -571.70 kJ

ΔSºsurr = - ΔHº / T = - (-571.70 kJ/mol) / (25 + 273.15 K) = 1917.0 J/K

More Calculation Examples Problem: The Overall reaction for the corrosion (rusting) of iron by oxygen is: 4Fe(s) + 3O2(g) → 2Fe2O3 (s). Use the following data to calculate the equilibrium constant at 25 °C. Substance ΔHfº (kJ/mol) Sº (J/K mol)

Fe2O3 (s) -826 90 Fe(s) 0 27 O2(g) 0 205

ΔHº = 2 x (-826) – 3 x (0.0) – 4x (0.0) = -1652 kJ ΔSº = 2 x (90) – 3 x (205) – 4x (27) = -543 J/K

ΔGº=-1652 kJ – (25+273 K) x (-543 J/K) = -1.49 x 106 J -1.49 x 106 J = - (8.314 J/Kmol) (298K) ln K

K = e601

ΔG° = –RT ln K ΔG° = ΔH° – TΔS°

ΔH° = ΣnproductsΔHf° products – ΣnreactantsΔHf° reactants

ΔS°reaction = ΣnpS°products – ΣnrS°reactants

More Calculation Examples Problem: Consider the ammonia synthesis reaction N2 (g) + 3H2(g) ⇌ 2NH3(g) where ΔG° = -33.3 kJ/mol of N2 consumed at 25 °C. For each of the following mixtures of reactants and products at 25 °C, predict the direction in which the system will shift to reach equilibrium. a). Pammonia=1.00 atm, Pnitrogen=1.47atm, Phydrogen=0.01atm b). Pammonia=1.00 atm, Pnitrogen=1.00atm, Phydrogen=1.00atm

To predict the reaction direction, we can calculate the free energy change for the given condition. a) ΔG = ΔG° + RT lnQ Q = (Pammonia)2 / (Pnitrogen)(Phydrogen)3 = (1.00)2 / (1.47) (0.01)3 = 6.80 x 105

ΔG = -33.3 x 103 J/mol + 8.314 J/Kmol x (25+273) x ln(6.80 x 105) = 0 Since ΔG = 0, the system is already at equilibrium, and no shift will occur. b) The partial pressures given are all 1.00 atm, which means that the system is in the standard condition, that is, ΔG = ΔG° = -33.3 kJ/mol. Since ΔG < 0, the system will move to the right to reach equilibrium.

Summary Spontaneous Processes and Entropy The Second Law of Thermodynamics The Effect of Temperature on Spontaneity Free Energy Thermodynamics in Chemical Reactions The Dependence of Free Energy on Pressure Free Energy and Equilibrium